/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO

After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 },
it remains to prove termination of the 3-rule system
{ 0 ⟶ ,
  0 1 ⟶ 1 2 0 0 ,
  2 2 2 ⟶ 0 1 }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 3-rule system
{ 0 ⟶ ,
  1 0 ⟶ 0 0 2 1 ,
  2 2 2 ⟶ 1 0 }

Loop of length 26 starting with a string of length 8
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.ba.aaaaaa
	rule ba-> aacb at position 0
.aacb.aaaaaa
	rule ba-> aacb at position 3
.aacaacb.aaaaa
	rule a->  at position 3
.aacacb.aaaaa
	rule a->  at position 3
.aaccb.aaaaa
	rule ba-> aacb at position 4
.aaccaacb.aaaa
	rule a->  at position 4
.aaccacb.aaaa
	rule a->  at position 4
.aacccb.aaaa
	rule ccc-> ba at position 2
.aabab.aaaa
	rule ba-> aacb at position 4
.aabaaacb.aaa
	rule ba-> aacb at position 7
.aabaaacaacb.aa
	rule a->  at position 7
.aabaaacacb.aa
	rule a->  at position 7
.aabaaaccb.aa
	rule ba-> aacb at position 8
.aabaaaccaacb.a
	rule a->  at position 8
.aabaaaccacb.a
	rule a->  at position 8
.aabaaacccb.a
	rule ccc-> ba at position 6
.aabaaabab.a
	rule ba-> aacb at position 6
.aabaaaaacbb.a
	rule ba-> aacb at position 10
.aabaaaaacbaacb.
	rule ba-> aacb at position 9
.aabaaaaacaacbacb.
	rule a->  at position 9
.aabaaaaacacbacb.
	rule a->  at position 9
.aabaaaaaccbacb.
	rule ba-> aacb at position 10
.aabaaaaaccaacbcb.
	rule a->  at position 10
.aabaaaaaccacbcb.
	rule a->  at position 10
.aabaaaaacccbcb.
	rule ccc-> ba at position 8
.aabaaaaababcb.
	rule ba-> aacb at position 8
.aabaaaaaaacbbcb.